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Valid Covariance Matrices

  1. Nov 30, 2014 #1
    I'm trying to understand what makes a valid covariance matrix valid. Wikipedia tells me all covariance matrices are positive semidefinite (and, in fact, they're positive definite unless one signal is an exact linear combination of others). I don't have a very good idea of what this means in practice.

    For example, let's assume I have a real-valued covariance matrix of the form:

    [tex]\mathbf{R}=\left[
    \begin{array}{ccc}
    1 & 0.7 & x \\
    0.7 & 1 & -0.5 \\
    x & -0.5 & 1
    \end{array}
    \right][/tex]
    where [itex]x[/itex] is some real number. What range of values can [itex]x[/itex] take?

    I can sort of see that [itex]x[/itex] is constrained by the other numbers. Like it can't have magnitude more than 1, because the diagonals are all 1. However, it is also constrained by the off-diagonals.

    Of course, for my simple example, I can solve the eigenvalue problem for eigenvalues of zero to give me the range of values (roughly -0.968465844 to 0.268465844)... but this hasn't really given me any insight in a general sense.

    I feel like there might be a neat geometrical interpretation that would make this obvious.

    Can anyone offer any insight?
     
    Last edited: Nov 30, 2014
  2. jcsd
  3. Dec 1, 2014 #2

    mathman

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    Gold Member

    I don't know if this a complete answer. However assume you have three random variables X, Y, Z each with variance 1, cov(X,Y) = 0.7, cov(X,Z) = x, and cov(Y,Z) = -0.5. For simplicity assume all means = 0.
    Consider E((X±Y±Z)2)≥0 for all possible sign combinations. This will give you four bounds on x. This may be the best, although I am not sure.
     
  4. Dec 2, 2014 #3

    Stephen Tashi

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    The terminology "covariance matrix" is ambiguous. There is a covariance matrix for random variables and there is a covariance matrix computed from samples of random variables. I don't think it works to claim that the sample covariance matrix is just the covariance matrix of a population consisting of the sample because the usual way to compute the sample covariance involves using denominators of n-1 instead of n.
     
  5. Dec 8, 2014 #4
    I'll have to give this some thought. It's not obvious to me how this works.
     
  6. Dec 8, 2014 #5
    What's the difference between these two? Do either/both have to be Hermitian positive semidefinite? That's the sort I'm interested in.
     
  7. Dec 9, 2014 #6

    mathman

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    1.5+cov(X,Y)+cov(X,Z)+cov(Y,Z)≥0
    1.5-cov(X,Y)+cov(X,Z)-cov(Y,Z)≥0
    1.5+cov(X,Y)-cov(X,Z)-cov(Y,Z)≥0
    1.5-cov(X,Y)-cov(X,Z)+cov(Y,Z)≥0
     
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